Perfect Squares.....

To begin with, let us note that the number we are looking at (namely $AABB$ ) is a number divisible by $11$ , since $A-A+B-B=0\equiv 0 \pmod{11}$ . You may find further insight into the divisibility rules by $11$ $\href{http://www.mathsisfun.com/divisibility-rules.html}{here}$ .

Let us say that $AABB=M^{2}$ . Since $AABB\equiv 0 \pmod{11}$ , note that $AABB$ can be factorized as $AABB=11^{n}\cdot R$ , where $n\in \mathbb{N}$ and $R$ represents the other factors of $AABB$ .

Consequently, we see that $M^{2}$ must also be divisible by $11$ (since $M^{2}=11^{n}\cdot R$ ). Let us also note that if a product of integers is divisible by a particular integer, then that must mean that at least one of the multipliers in the product must also be divisible with that integer. In this case, it is obvious that $11|M$ .

Therefore, we see that $M$ is a multiple of $11$ ! This actually does all the work for us, considering that there are only $7$ multiples of $11$ that form $4$ -digit numbers and those are $33,44,55,66,77,88$ and $99$ .

At this point, we may easily fine the squares for each of them, which are $1089,1936,3025,4356,5929,7744$ and $9801$ accordingly. From this, we can see that $7744$ is the only number satisfying this condition, which is ironically the square of $88$ and also the example given to us by Tanmay.

Thus, the answer is $\boxed{1}$ .

AABB= 1000A+100A+10B+B=1100A+11B=11(100A+B) Now since AABB=k^2 therefore 100A+B=11.m^2 as it is a perfect square. Since m=1,2,3,4,5,6,7,8,9 ( not 10 as A and B are single digit integers) Thus 100A+B=11,44,275,396,539,704,891 Since we will only get the single positive integer values of A and B in 1 case only. Therefore the answer is 1